Linear Approximation

Linear approximation says: near a point, a differentiable function behaves like its tangent line. The gap between the function and the tangent line shrinks faster than the displacement itself.

Prerequisites

• Derivatives, Differentials, and the Chain Rule --- definition of $f^{\prime }(x)$
• The Mean Value Theorem --- needed for the stronger error bound

Notation

• $f(x)$ --- a differentiable function
• $h$ --- a small displacement from point $a$ (also written $\Delta x$)
• $L(x)=f(a)+f^{\prime }(a)(x-a)$ --- the tangent line at $a$
• $R(h)$ --- the remainder (error) of the approximation

Statement

The tangent line $L(x)$ at $a$ approximates $f(x)$ near $a$. The vertical gap (error $R(h)$) shrinks faster than $h$ itself.

Hypothesis: $f$ is differentiable at $a$.

Thesis: $f(a+h)=f(a)+f^{\prime }(a)\cdot h+R(h)$, where $\underset{h\to 0}{\lim }\frac{R(h)}{h}=0$.

The notation $R(h)=o(h)$ means “the error is negligible compared to $h$.” Concretely:

$f(a+h)\approx f(a)+f^{\prime }(a)\cdot h$

with an error that vanishes faster than $h$.

Proof

Step 1. The derivative is defined as:

$f^{\prime }(a)={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$

Step 2. Define the remainder $R(h)=f(a+h)-f(a)-f^{\prime }(a)\cdot h$.

Step 3. Divide by $h$:

$\frac{R(h)}{h}=\frac{f(a+h)-f(a)}{h}-f^{\prime }(a)$

Step 4. Take $h\to 0$. By definition of derivative (Step 1), the right side $\to 0$.

${\lim }_{h\to 0}\frac{R(h)}{h}=0\square$

The stronger error bound: $R(h)=O(h^2)$

The proof above only requires differentiability at $a$ and gives $R(h)=o(h)$. With a stronger hypothesis, we get a quantitative bound:

Hypothesis: $f$ is twice differentiable on an interval containing $a$ and $a+h$, with $|f^{\prime \prime }|\le M$ on that interval.

Thesis: $|R(h)|\le \frac{M}{2}{h}^2$

Proof

Step 1. By Taylor’s theorem with remainder, there exists $c$ between $a$ and $a+h$ such that:

$R(h)=\frac{f^{\prime \prime }(c)}{2}\cdot h^2$

Step 2. Take absolute values and bound $|f^{\prime \prime }(c)|\le M$:

$|R(h)|\le \frac{M}{2}\cdot h^2\square$

Taylor’s remainder theorem is proved via the Mean Value Theorem applied to a helper function.

Why this matters for the FTC

In the FTC Part 2 proof, Step 3 uses linear approximation on each subinterval:

$F(x_{i+1})-F(x_i)\approx F^{\prime }(x_i)\cdot \Delta x$

The error in each term is $|R_i|\le \frac{M}{2}\Delta x^2$ (where $M=\max |F^{\prime \prime }|$). Summing $n$ terms:

$\text{total error}\le n\cdot \frac{M}{2}\Delta x^2=\frac{M}{2}\cdot (b-a)\cdot \Delta x\to 0$

since $n\cdot \Delta x=b-a$ is constant while $\Delta x\to 0$.

Without the $O(h^2)$ bound, each error is only $o(\Delta x)$, and summing $n$ terms that are each $o(\Delta x)$ does not obviously give something small (uniform convergence arguments are needed). The $O(h^2)$ bound avoids this: $n\cdot O(\Delta x^2)=O(\Delta x)\to 0$.

Logical dependency

The $O(h^2)$ bound uses MVT. The FTC implies MVT as a consequence. This is not circular:

1. The FTC proof only needs the weaker $o(\Delta x)$ result (which uses no MVT) if you handle uniform continuity carefully. The $O(h^2)$ bound is a convenience.
2. The logical chain is: definition of derivative $\to$ MVT $\to$ $O(h^2)$ error bound $\to$ FTC Part 2. Each step uses only what came before. “FTC is stronger than MVT” means FTC implies MVT — not that FTC is proved without MVT.

See also

• Derivatives, Differentials, and the Chain Rule --- the differential $df=f^{\prime }(x)dx$ is linear approximation in differential notation
• The Mean Value Theorem --- needed for the $O(h^2)$ error bound
• The Fundamental Theorem of Calculus --- uses linear approximation in its proof